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From: whygee on 14 Apr 2010 17:47
George Jefferson wrote:
> I'*M(int(I)) + I/C = 0
> I'*[Q/C - V]/I + I/C = 0
> I*I'*[Q/C - V] + I^2/C = 0
> Q - V*C + I/I' = 0
> diff,
> I + [I'^2 - I*I'']/I'^2 = 0
> I*I'^2 + I'^2 - I*I'' = 0
> [I + 1]*I'^2 = I*I''
> > If you plot this de out you should get the solution if I didn't make any
> mistakes(We lost C but it will come back).

I was half-expecting that you come up with e=mc^2
or something like that ;-)

yg
--
http://ygdes.com / http://yasep.org
From: George Jefferson on 14 Apr 2010 21:13


"whygee" <yg(a)yg.yg> wrote in message news:hq5fci$615$1(a)speranza.aioe.org...
> George Jefferson wrote:
>> I'*M(int(I)) + I/C = 0
>> I'*[Q/C - V]/I + I/C = 0
>> I*I'*[Q/C - V] + I^2/C = 0
>> Q - V*C + I/I' = 0
>> diff,
>> I + [I'^2 - I*I'']/I'^2 = 0
>> I*I'^2 + I'^2 - I*I'' = 0
>> [I + 1]*I'^2 = I*I''
>> > If you plot this de out you should get the solution if I didn't make
>> > any
>> mistakes(We lost C but it will come back).
>
> I was half-expecting that you come up with e=mc^2
> or something like that ;-)
>

Well, I overcomplicated things

It's much easier to see that

V - R(Q)*I - Q/C = 0

and I = Q'

So we have

V - R(Q)*Q' - Q/C = 0

which is the nonlinear DE if R(Q) is nonlinear. Knowing the expression for R
lets us solve for Q then we can solve for I.


From: George Jefferson on 14 Apr 2010 21:44
Here is a matlab script that simulates an M-C circuit:

Script - memristor

options = odeset('RelTol',1e-5,'AbsTol', 1e-5, 'MaxStep', 0.001);
[T,Q] = ode113(@memcap, [0 5], 0, options);

hold on;
plot(T, Q);
plot(T, [diff(Q)./diff(T); 0], 'Color', 'red');
hold off;

-------------
function - memcap

function dQdt = memcap(t, Q)

V = 1;
C = 1;

dQdt = (V - Q/C)./Res(Q);

function r = Res(Q)
r = 1 + 0.25*cos(10*Q);
---------------

Res(Q) = M(Q). If M(Q) is constant you get the standard RC circuit.



From: George Jefferson on 14 Apr 2010 22:22
http://i39.tinypic.com/2uhsvvo.jpg

Showing a non-linear resistor for various parameter adjustments(similar to
the code I posted).

We you can see they are all approximately "resistor like". Remember that it
comes from an RC circuit with R depending on charge.

Because the charge will eventually reach a maximum value(C*V for this
circuit) if the resistance never becomes negative, the resistance must
eventually become constant. i.e., The cap will eventuallly max out and the
current will decrease causing a steady steady state in the resistance.

Again, the max charge that can pass through the resistor is C*V which is the
max charge the cap can hold. Because there is a maximum charge R(Q)->R(C*V)
and C*V is a constant, hence R(Q)->constant and which is why the curves seem
to become more and more like and R-C circuit as time goes on.

If we replaced the cap with a linear resistor then the charge Q will
approach infinity and R(Q) could be drift off. There are a few possibilities
that could happen.

What we sorta need to know is if R(Q) is generally increasing, constant, or
decreasing as Q->oo. Each one implies a different steady state for
different circuits.

I imagine that for true memristors there is no singularities since this
would create some impossibile physical situations.




From: Robert Baer on 15 Apr 2010 01:02
Bill Sloman wrote:
> On Apr 14, 7:51 am, Robert Baer <robertb...(a)localnet.com> wrote:
>> Jan Panteltje wrote:
>>> On a sunny day (Tue, 13 Apr 2010 11:24:33 -0700 (PDT)) it happened Wanderer
>>> <wande...(a)dialup4less.com> wrote in
>>> <b74e200b-933e-4848-a9b4-a9a66a623...(a)f17g2000vbd.googlegroups.com>:
>>>> HP found a way to make memristors
>>>> http://spectrum.ieee.org/semiconductors/design/the-mysterious-memristor
>>>> and I'm trying to make heads or tails of them. If they came in 0805
>>>> packages and you could buy them from digi-key, how would you spec
>>>> them? They say it has the units of ohms but if its a constant its just
>>>> a resistor, so you can't spec a 3 ohm memristor. Can you spec a 3
>>>> Weber/coulomb memristor? Doing unit analysis on the equations
>>>> R = dv/di , C = dq/dv, L = dphi/di, M = dphi/dq, V = dphi/dt, I = dq/
>>>> dt
>>>> I get
>>>> RC = M/L = idt/di has units of time
>>>> L/R = CM = vdt/dv has units of time
>>>> LC = (idt)(vdt)/(dvdi) has units of time squared
>>>> R/M (idtdv)/(vdtdi) is unit less
>>>> Does this thing break traditional circuit analysis?
>>> I did a read very simple explanation:
>>> The electrical current moves some atoms in a grid.
>>> that changes the resistance permanently.
>>> Reversing the current moves the atoms back.
>>> This can be done very fast (much faster then programming FLASH).
>>> I am sure that the chips that will be marketed will have a controller build in,
>>> and you will just be able to interface with it in the usual way.
>> ..something like a Coulomb meter or one of the definitions of current?
>> Like the amount of silver plated per unit of time?
>> And this is claimed to be new?
>
> Do pay attention.
>
> The apparent "resistance" of a Coulomb meter doesn't change with the
> amount of metal that has been plated out; the memistor has a "memory"
> and its "resistance" reflects its history.
>
> --
> Bill Sloman, Nijmegen
Are you trying to tell me that the silver plating cell originally
devised to measure / be a current standard does _not_ change its
resistance as the silver gets plated / unplated?
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